derivative (1/(y^2)-7/(y^4))(y+9y^3)

Lösung

ableitung von

(\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) (y + 9 y^{3})

\frac{62}{y ^{2}} + \frac{21}{y ^{4}} + 9

Schritte zur Lösung

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) (y + 9 y^{3}))

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = \frac{1}{y ^{2}} - \frac{7}{y ^{4}}, g = y + 9 y^{3}

= \frac{d}{d y} (\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) (y + 9 y^{3}) + \frac{d}{d y} (y + 9 y^{3}) (\frac{1}{y ^{2}} - \frac{7}{y ^{4}})

\frac{d}{d y} (\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) = - \frac{2}{y ^{3}} + \frac{28}{y ^{5}}

\frac{d}{d y} (y + 9 y^{3}) = 1 + 27 y^{2}

= (- \frac{2}{y ^{3}} + \frac{28}{y ^{5}}) (y + 9 y^{3}) + (1 + 27 y^{2}) (\frac{1}{y ^{2}} - \frac{7}{y ^{4}})

Vereinfache

(- \frac{2}{y ^{3}} + \frac{28}{y ^{5}}) (y + 9 y^{3}) + (1 + 27 y^{2}) (\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) : \frac{62}{y ^{2}} + \frac{21}{y ^{4}} + 9

= \frac{62}{y ^{2}} + \frac{21}{y ^{4}} + 9

x \to 3 lim (2 x - 3)

2 y^{^{''}} + 33 y = 0

x \to 1 lim (3 x)

k = 1 \sum \infty \frac{( - 1 ) ^{k}}{k}

\frac{d}{d x} ((6 x + 5)^{1.5})